TITLE: Birational and faithfully flat $\implies$ isomorphism
QUESTION [16 upvotes]: Let $A \subseteq B$ be integral domains with the same field of fractions. Assume that $A \to B$ is faithfully flat. Why do we have $A=B$?
This is an exercise in Matsumura's book. Here is my idea: If $b \in B$, consider $I = \{a \in A : ab \in A\}$. This is an ideal of $A$. By asumption $I \neq 0$, and our goal is to show that $I=A$. It suffices to prove $IB=B$. But how can we achieve this?

REPLY [11 votes]: We can show $I = A$ as follows, the inclusion $I \subseteq A$ being clear. Choose $a \in A$ and write $b = a_1/a_2$. Then $aba_2 = a\cdot a_1 \in A$ and so $a\cdot a_1 \in a a_2 B \cap A = aa_2 A$ where the last equality comes from faithful flatness. It follows that $aba_2 \in aa_2A$ and so $aba_2 = a a_2 a'$ for some $a' \in A$. Thus $ab = a a' \in A$ and so indeed $a \in I$. Thus $I = A$.